Spring 17
What Is A Parametric Curve? y P(x, y) x 1. Let a point P on a curve have Cartesian coordinates (x, y). We can think of the curve as being traced out as the point P moves along it. 3. In this way we can think of both x and y as functions of t. 4. x f (t) and y g(t) are called parametric equations of the curve; t is called a parameter. Example 1 Let { x cos t y sin t with t π. These are parametric equations of a circle: x + y cos t + sin t 1. Figure: The circle x + y 1 In terms of t, the circle is being traced out counter clockwise.
Example Let { x t 3 y t with t R. We can eliminate the parameter: t x 1/3 y We could also say ( x 1/3) x /3. Figure: The curve y 3 x y 3 t 6 x y 3 x. Observe: t 1 x 1, y 1; t x 8, y 4; etc. Example 3 Let { x t y t 3 3t with t R. For this example, you could eliminate the parameter, but it would be messy. For instance, you would have to consider two cases: t ± x. Generally, parametric curves are most suitable for relations in which no single function can easily describe the curve.
Most General Form of a Curve Parametric curves are the most general type of curve. The function y f (x) can be described parametrically by { x t y f (t), t a parameter In essence, x is the parameter of a curve described by the function y f (x). All curves we shall look at later, polar curves and curves in 3 dimensions, can be described parametrically. Derivatives of Parametric Curves To do calculus with parametric curves we will need formulas for derivatives in terms of the parameter. Suppose x and y are functions of a parameter t. Then As always, the derivative will be the slope of the curve at the point (x, y); the difference is that everything will be calculated in terms of the parameter t.
Example 4; Example Revisited We have x t 3, y t ; so t 3t 3t, t. This is the same answer you get if you eliminate the parameter: t x 1/3 y x /3 3 x 1/3. Example 5; Example 3 Revisited We have x t, y t 3 3t; so 3t 3. t The critical points are at t ±1; t. The coordinates of the critical points are (, ), (1, ±).
The Second Derivative in Parametric Form d y d ( ), with y From Examples 4 and : x t 3, y t ; 3t. So d y d ( ) 3t 3 3t 3t 9t 4. Or, since t x 1/3 and y x /3 : 3x 1/3 d y 9x 4/3.. Example 6; Examples 3 and 5 Revisited In Examples 3 and 5 we had: x t, y t 3 3t, and Hence d y Thus d ( ) 3t 3 t d ( 3t 3 ) t 3t 3 t. ( 3 + 3 ) t d y > t > and d y < t <. So (, ) is an inflection point. (Check the graph!) t 3 4 t + 1 t 3.
Example 7: The Cycloid Consider the parametric curve with x t sin t; y 1 cos t. We shall do a typical analysis of the first two derivatives to figure out how the graph of this curve, which is called a cycloid, looks: sin t 1 cos t and d y cos t cos t sin t 1 (1 cos t) 3 (1 cos t), as you may check. Since the second derivative is non-positive, the graph will always be concave down. There are two types of critical points: 1. vertical tangents, when cos t 1 : namely at t, ±π, ±4π,.... So (x, y) (kπ, ), k Z.. horizontal tangents, when sin t but cos t 1; namely at t ±π, ±3π,.... So (x, y) ((k + 1)π, ), k Z. Graph of the Cycloid, for t 4π 1. The graph is increasing if > sin t > < t < π, π < t < 3π. The graph is decreasing if < sin t < π < t < π, 3π < t < 4π Figure: A cycloid.
Integral Formulas in Parametric Form Every integral formula that you are familiar with from Chapter 6: 1. area, eg. A b a y. volume, eg. V b a πy or V b a πx y 3. length, eg. L b a 1 + ( ) 4. surface area, eg. SA ( ) b a 1 πx + ; can be applied to parametric curves by substituting for x and y in terms of the parameter t, and then simplifying. In other words, these problems are just fancy change of variable problems. But, you must be careful in changing the limits of integration as you make the substitutions. Your new integral will be in terms of t, and so should the limits of integration. Example 8: Area of a Circle Circle of radius a has parametric equations x a cos t, y a sin t. a A 4 y 4 π/ a sin t ( a sin t) π/ 4a sin t π/ ( ) 1 cos(t) 4a [ a t sin(t) ] π/ πa
Example 9: Volume of a Sphere Rotate the quarter circle around the x-axis, and multiply by two. a V πy π π/ πa 3 a sin t ( a sin t) π/ (1 cos t)( sin t) 1 πa 3 (1 u ) du, with u cos t πa 3 [ u 1 3 u3 ] 1 4 3 πa3 Example 1: Area of the Ellipse x a + y b 1 Parametric equations of an ellipse: x a cos t, y b sin t. a A 4 y 4 4ab 4ab ab π/ π/ b sin t ( a sin t) π/ sin t ( ) 1 cos(t) [ t sin(t) ] π/ πab
Example 11: Volume of a Torus A torus is obtained by rotating a circle, x b + a cos t, y a sin t, for a < b, around the y-axis. b+a V 4π b a π πx y (b + a cos t) a sin t ( a sin t) π 4πa (b sin t + a cos t sin t) π ( ) 1 cos(t) 4πa b + [ πa b t sin(t) ] π π a b Example 1 Find the area within the loop of the curve x t, y t 3 3t. 3 A 3 3 y, if y > (t 3 3t) t (6t t 4 ) [ t 3 5 t5 ] 3 4 3 5
What Are Polar Coordinates? y P(x, y) r θ O x 1. Let the point P have Cartesian coordinates (x, y). Let r be the length of OP. 3. Let θ be the angle the line OP makes with the positive x-axis. 4. Then r and θ are called the polar coordinates of P. We have Equivalently, x r cos θ and y r sin θ. r x + y and tan θ y x. The value of θ depends on which quadrant (x, y) is in. Example 1 Find the polar coordinates of the following points. Warning: polar coordinates are not unique because θ is not unique. (x, y) (r, θ) with angles chosen from [ π, π] (, ) r, θ or θ π (1, 1) r, θ π/4 (, 1) r 1, θ π/ (, 3) r 3, θ π/ or θ 3π/ (, ) r, θ π or θ π ( 1, 3) r, θ 4π/3 or θ π/3 (3, 4) r 9 + 16 5, θ tan 1 ( 4 3).97 radians (, ) r, θ can be anything.
Negative Values of r According to the definition of polar coordinates, r should be positive. However, we can extend the definition to include negative values of r. Suppose r <. P( x, y) y Q(x, y) r O θ x 1. Let Q have coordinates (x, y) and polar coordinates ( r, θ).. Then the point P with polar coordinates (r, θ) is defined to be the point with polar coordinates ( r, θ ± π). 3. That is, OP OQ. Example The point with polar coordinates ( (r, θ) 3, π ) is the same as the point with polar coordinates (3, π ) ( π 3, π ). Its Cartesian coordinates are (x, y) (, 3). y (, 3) π/ π/ (, 3) x
Polar Curves If r f (θ), the polar curve of f is the set of all points whose polar coordinates (r, θ) satisfy the equation r f (θ). There is a difference between the graph of r f (θ) and the polar curve of the equation r f (θ). The former is the set of points (x, y) such that x θ and y f (θ). These graphs are nothing new; exactly like all the previous examples you ve done in Chapters 3 and 4. The polar curve is something new. On the polar curve, the coordinates are (x, y) (f (θ) cos θ, f (θ) sin θ). Example 3: r 1 Figure: Graph of r 1 Figure: Polar curve: r 1
Example 4: r sin θ Figure: Graph of r sin θ Figure: Polar curve: r sin θ Example 4, Continued You can convince yourself that the polar curve is a circle by changing the polar equation into a Cartesian equation: r sin θ r r sin θ x + y y x + y y x + y y + 1 1 x + (y 1) 1 This last equation is the equation of a circle with centre (, 1) and radius 1. Warning: it won t always be this easy to switch from the equation of a polar curve to its Cartesian equation! In fact, we will rarely do it.
Exampl 5: r 1 + cos θ Figure: Graph of the function r 1 + cos θ Figure: Cardioid with polar equation: r 1 + cos θ Example 6: r + cos θ Figure: Graph of the function r + cos θ Figure: Curve with polar equation: r + cos θ
Example 7: r 1 cos θ Figure: Graph of the function r 1 cos θ Figure: Limaçon with polar equation: r 1 cos θ Example 7, Continued The limaçon has an interesting feature: a loop within a loop. For which angles does the curve of r 1 cos θ trace out the outer loop? the inner loop? The two loops have only one point in common, the origin (, ), at which r. The boundary angles of the loops are found by solving r cos θ 1 θ π 3 or 5π 3. The outer loop, when r >, is traced for π 3 < θ < 5π 3. For π 3 < θ < π, r < and the curve traces out the inner loop. 3
Example 8: r sin(θ) Figure: Graph of the function r sin(θ) Figure: Four leaved rose with polar equation: r sin(θ) Example 9: r sin(3θ) Figure: Graph of the function r sin(3θ) Figure: Three leaved rose with polar equation: r sin(3θ)
Example 1: r sin θ + cos θ Figure: Graph of the function r sin θ + cos θ Figure: Circle with polar equation: r sin θ + cos θ Example 1, Continued We can confirm that the polar curve of r sin θ + cos θ is a circle, by finding its Cartesian equation. r sin θ + cos θ r r sin θ + r cos θ x + y y + x So the circle has centre ( 1, 1 ) and radius 1/. x x + 1 4 + y y + 1 4 1 4 + 1 4 ( x 1 ( + y ) 1 ) 1
Example 11: r θ/5 Figure: Graph of the function r θ/5, θ Figure: Archimedean spiral with polar equation: r θ/5, θ Example 1: r e θ/5 Figure: Graph of the function r e θ/5 Figure: Logarithmic spiral with polar equation: r e θ/5
Derivatives for Polar Curves Since the parametric equations for a polar curve are x r cos θ, y r sin θ, the first derivative of a polar curve is dθ dθ sin θ dr dθ + r cos θ cos θ dr. dθ r sin θ The formula for the second derivative is so messy we won t even try to write it down! Example 1 Find the slope of the tangent line to the circle with polar equation r 4 cos θ at the point with θ π/6. sin θ dr dθ + r cos θ cos θ dr dθ r sin θ 4 sin θ + 4 cos θ 4 cos θ sin θ 4 cos θ sin θ sin θ cos θ cot(θ) sin θ cos θ So at θ π/6 the slope is cot(π/3) 1/ 3.
Example : Tangent Lines to Polar Curves at the Origin If r for θ α, and dr/dθ at θ α, then sin α dr dθ + θα θα cos α dr dθ θα tan α. That is, the line θ α is tangent to the curve at the origin. The figure to the right illustrates the example r sin 3θ for which the tangent lines at the origin have equations θ, π/3 or π/3. Example 3: Critical Points of a Cardioid Looking at the graph of the cardioid r 1 + cos θ, we expect to find six critical points: namely where the graph has its three vertical tangents and its three horizontal tangents. The parametric equations of the cardioid are so x r cos θ; y r sin θ { x (1 + cos θ) cos θ y (1 + cos θ) sin θ
Horizontal Tangents to the Cardioid r 1 + cos θ dθ d ((1 + cos θ) sin θ) dθ sin θ + cos θ + cos θ cos θ + cos θ (1 cos θ) cos θ + cos θ 1 ( cos θ 1)(cos θ + 1) cos θ 1 or cos θ 1 Find the critical points: cos θ 1 r (x, y) (, ). cos θ 1 r 3 ( ) 3 and sin θ ± 3 3 (x, y) 4, ±3. 4 Vertical Tangents to the Cardioid r 1 + cos θ dθ d ((1 + cos θ) cos θ) dθ sin θ cos θ sin θ sin θ cos θ sin θ(1 + cos θ) cos θ 1 or sin θ Find the critical points: cos θ 1 r 1 and sin θ ± 3 (x, y) ( 1 4, ± 3 4 ). sin θ cos θ ±1 r or. Take (x, y) (, ).
Example 4: Logarithmic Spiral r e θ/5 Revisited Note: for the following, the algebra is easier if r e θ, but then the curve spirals out so quickly that it is hard to get a suitable graph. The parametric equations of this spiral are x e θ/5 cos θ, y e θ/5 sin θ; 1/5eθ/5 sin θ + e θ/5 cos θ 1/5e θ/5 cos θ e θ/5 sin θ sin θ + 5 cos θ cos θ 5 sin θ m sin θ + 5 cos θ cos θ 5 sin θ m sin θ + 5 cos θ m(cos θ 5 sin θ) (1 + 5m) sin θ (m 5) cos θ tan θ m 5 1 + 5m 1. m tan θ 5. This means that all the points on the spiral with slope zero lineup along the line y 5x.. m ± tan θ 1 5. This means that all the points on the spiral with undefined slope lineup along the line y x/5. 3. m 1 tan θ 1 13. This means that all the points on the spiral with slope 1 lineup along the line y 1 13 x. 4. m 1 tan θ 13 1. This means that all the points on the spiral with slope -1 lineup along the line y 13 1 x.
Area within a Polar Curve θ β y r f (θ) θ α x The area within the polar curve with equation r f (θ), between the rays θ α and θ β, is given by A 1 β α r dθ. Proof: see the book. It s not hard: use partitions of the interval [α, β] and Riemann sums to approximate the area. You need to know that the area of a circle sector is given by 1 r θ. Example 5: Area In A Cardioid Figure: Cardioid with polar equation: r 1 + cos θ By symmetry, we can double the area of the top half. So A ( 1 π ) r dθ π π π (1 + cos θ) dθ (1 + cos θ + cos θ) dθ ( 3 + cos θ + 1 ) cos(θ) dθ [ ] 3 sin(θ) π θ + sin θ + 3π 4
Example 6 Fiind the area of one petal of the three-leaved rose r sin(3θ). r 3θ or π θ or π 3. Figure: r sin(3θ) A 1 π/3 π/3 4 sin (3θ) dθ (1 cos(6θ)) dθ [θ 16 sin(6θ) ] π/3 π 3 Example 7 Find the area inside the circle with polar equation r sin θ + cos θ. r sin θ + cos θ tan θ 1 θ π 4 or 3π 4 Figure: r sin θ + cos θ A 1 1 1 3π/4 π/4 3π/4 π/4 (sin θ + cos θ) dθ (1 + sin θ cos θ) dθ [ θ + sin θ ] 3π/4 π/4 π
Example 8 Find the area within each of the inner and outer loops of the limaçon: Recall, that the outer loop, r >, is traced out for θ [π/3, 5π/3], and that inner loop, r <, is traced out for θ [ π/3, π/3]. Figure: r 1 cos θ Let A i be the area of the inner loop; let A o be the area of the outer loop. Example 8, Continued A ( 1 π π/3 π π/3 (1 cos θ) dθ ) π π/3 (1 4 cos θ + 4 cos θ) dθ (3 4 cos θ + cos(θ)) dθ [3θ 4 sin θ + sin(θ)] π π/3 3π 3(π/3) + 4 sin(π/3) sin(π/3) π + 3 3 A i ( 1 π/3 (1 cos θ) dθ ) [3θ 4 sin θ + sin(θ)] π/3 3(π/3) 4 sin(π/3) + sin(π/3) π 3 3
Area Between Curves y θ β r 1 θ α r x If the two polar curves r 1 and r intersect at θ α and θ β, then the area between the curves is given by A 1 β α ( r 1 r ) dθ. This is simply the area within r 1 minus the area within r. Example 9 Find the area inside the limaçon r 1 + cos θ but outside the circle r. Have: 1 + cos θ cos θ 1/ θ ±π/3. A ( 1 π/3 π/3 π/3 ( (1 + cos θ) ) dθ ( 3 + 4 cos θ + 4 cos θ ) dθ ( 1 + 4 cos θ + cos(θ)) dθ [ θ + 4 sin θ + sin(θ)] π/3 π/3 + 4 sin(π/3) + sin(π/3) 5 3/ π/3 )
Example 1: A Very Tricky Example Find the area inside the circle r 3 cos θ and inside the circle r sin θ. Have: 3 cos θ sin θ tan θ 3 θ π/3. There is another intersection point, at the origin. However, the two circles pass through the origin for different values of θ. For the circle r sin θ, r if θ. But for the circle r 3 cos θ, r if θ π/. This means that it will require two separate integrals to find the total area of the region common to these two circles. Example 1, Continued Let A 1 be the area inside r sin θ for θ [, π/3]; let A be the area inside r 3 cos θ for θ [π/3, π/]. A A 1 + A. A 1 1 π/3 4 sin θ dθ π/3 (1 cos(θ)) dθ [ θ sin(θ) ] π/3 π 3 3 4 A 1 π/ π/3 1 cos θ dθ 3 3 π/ π/3 (1 + cos(θ)) dθ [ θ + sin(θ) ] π/ π/3 π 3 4 3
Example 11 Consider the two cardioids r 1 ± sin θ. 1 + sin θ 1 sin θ sin θ θ or π The third intersection point is the origin: r 1 + sin θ if θ π/; r 1 sin θ if θ π/. So again, finding areas of regions determined by these two cardioids can be quite tricky. Example 11, Continued The area inside r 1 + sin θ but outside r 1 sin θ is given by 1 π ( (1 + sin θ) (1 sin θ) ) dθ 1 π 4 sin θ dθ [ cos θ] π 4 The area of the region common to both cardioids is given by ( ) 1 π/ π/ 4 (1 sin θ) dθ (1 sin θ + sin θ) dθ π/ (3 4 sin θ cos(θ)) dθ [3θ + 4 cos θ 1 ] π/ sin(θ) 3π 4
Example 1: The Problem of Intersection Points Consider the polar curves r 1 + sin θ and r 4 sin θ. The polar curve r 4 sin θ comes in two parts: r ± sin θ, for θ [, π]. From the graphs we expect four intersection points. Substituting for r from one equation into the other: (1 + sin θ) 4 sin θ 1 + sin θ + sin θ 4 sin θ 1 sin θ + sin θ (1 sin θ) sin θ 1 θ π/, at (x, y) (, ). Example 1: What are the other three intersection points? (, ) is one obvious other intersection point. For r 1 + sin θ, r if θ 3π/; for r 4 sin θ, r if θ or π. But how to get the other two intersection points? Recall that the point with polar coordinates (r, θ) is the same as the point with polar coordinates ( r, θ ± π). Rewrite the polar equation r 4 sin θ as ( r) 4 sin(θ ± π) r 4( sin θ) 4 sin θ, for π θ. Now substitute r 1 + sin θ : (1 + sin θ) 4 sin θ 1 + sin θ + sin θ 4 sin θ sin θ + 6 sin θ + 1 sin θ 3 ± sin θ 3 + θ 9.879 or 17.11 in degrees.
Example 1, Concluded If sin θ 3 +, then cos θ ± 1 ( 3 + ) ± 3 4, and r 1 + ( 3 + ) +. The Cartesian coordinates of the remaining two intersection points are: (x, y) (r cos θ, r sin θ) ( ±4( 1) 3 4, 14 1 ) (±.816147336,.141356) Example 13: Let r e θ/5, for 8π θ 1π. What is the area within this spiral chamber? For θ [, π], let r o e (1π+θ)/5, r i e (8π+θ)/5. Then r o ar i, for a e π/5, and ro ri a ri ri (a 1)ri (a 1) a 8 e θ/5. So A a 1 π (a 1) a 8 (a 1) a 8 (a 1) a 8 a 8 e θ/5 dθ [e θ/5] π 5 5 ( ) e 4π/5 1 5 (a 1) 19.9